if an object with mass m is dropped from rest, one model for its speed v after t seconds, taking air resistance into account is
v= mg/c(1-e^(-ct/m)) where g is the acceleration due to gravity and c is a positive constant describing air resistance.
a) calculate lim t->infinity (v.
answer: mg/c
----->what is the meaning of this limit
1)it is the time it takes the object to reach its maximum speed.
2)it is the speed the object approaches as time goes on.
3)it is the time it takes for the object to stop.
b) for fixed t us l'hospital's rule to calculate lim c->0^+ (v).
answer : tg
----->what can you conclude about the velocity of a falling object in a vacuum.
1)the velocity of a falling object in a vacuum is directly proportional to the amount of time it falls.
2)the heavier the object is the faster it will fall in a vacuum.
3)an object falling in a vaccuum will accelerate at a slower rate than an object not in a vacuum.
a) calculate lim t->infinity (v.
answer: mg/c
----->what is the meaning of this limit
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It is the "terminal velocity" (Google that) which is the speed where the drag force equals the weight. One uses a parachute so that one's terminal velocity is low enough to hit the ground without quite dying.
So
2)it is the speed the object approaches as time goes on.
AND
In a vacuum, c = 0 so terminal velocity is undefined.
a) The meaning of the limit lim t->infinity (v) is that it represents the speed that the object approaches as time goes on. In other words, it is the maximum speed that the object will reach after an infinite amount of time. This limit is equal to mg/c.
b) Based on the result of using l'Hospital's rule, lim c->0+ (v) evaluates to tg. This conclusion can be drawn about the velocity of a falling object in a vacuum: the velocity of a falling object in a vacuum is directly proportional to time. The heavier the object, the faster it will fall in a vacuum. The acceleration of an object falling in a vacuum is the same as an object not in a vacuum, as air resistance does not exist in a vacuum.
a) The meaning of the limit lim t->infinity (v) is the speed that the object approaches as time goes on. As t approaches infinity, the term e^(-ct/m) in the expression for v approaches zero. Therefore, the overall expression simplifies to v = mg/c. This means that as time goes on indefinitely, the object's speed will approach mg/c.
b) From the result of applying l'Hôpital's rule, which is lim c->0^+ (v) = tg, we can conclude that the velocity of a falling object in a vacuum is directly proportional to the acceleration due to gravity. As c approaches zero (representing a vacuum with no air resistance), the velocity formula simplifies to v = mg/t. This implies that in a vacuum, the velocity of a falling object is solely determined by the acceleration due to gravity, which is the same for all objects regardless of their mass.
a) The meaning of the limit lim t -> infinity (v) is that it represents the value that the speed v of the object approaches as time goes on. As t approaches infinity, the term e^(-ct/m) in the equation tends to zero, resulting in the numerator mg/c. Therefore, the limit represents the speed that the object approaches over an infinite amount of time, which is equal to mg/c.
So, option 2) It is the speed the object approaches as time goes on is the correct answer.
b) To calculate lim c -> 0+ (v) using L'Hospital's rule, we differentiate the numerator and denominator with respect to c to find their limits.
Taking the derivative of v with respect to c, we get:
dv/dc = -mg(e^(-ct/m)) / (c^2)(1 - e^(-ct/m))
Taking the limit of this derivative as c approaches 0, we get:
lim c -> 0+ (dv/dc) = -mg(1) / (0^2)(1 - 1) = -mg/0
Since we have an indeterminate form of -mg/0, we can use L'Hospital's rule again by differentiating the numerator and denominator with respect to c one more time.
Differentiating again, we get:
d^2v/dc^2 = -mg(e^(-ct/m))(c^2)(1 - e^(-ct/m)) + 2mg(e^(-ct/m))c(e^(-ct/m)) / (c^3)(1 - e^(-ct/m))^2
Taking the limit of this second derivative as c approaches 0, we get:
lim c -> 0+ (d^2v/dc^2) = -mg(1)(0)(1 - 1) + 2mg(1)(0)(0) / (0^3)(1 - 1)^2 = 0/0
Again, we have an indeterminate form of 0/0, so we can use L'Hospital's rule once more. By differentiating the numerator and denominator with respect to c a third time, we can find the limit.
Differentiating again, we get:
d^3v/dc^3 = 2mg(e^(-ct/m))(c^3)(1 - e^(-ct/m))^2 - 6mg(e^(-ct/m))c^2(e^(-ct/m))^2 - 2mg(e^(-ct/m))c^2(1 - e^(-ct/m))(e^(-ct/m)) / (c^4)(1 - e^(-ct/m))^3
Taking the limit of this third derivative as c approaches 0, we get:
lim c -> 0+ (d^3v/dc^3) = 2mg(0^3)(1 - 1)^2 - 6mg(0^2)(0)^2 - 2mg(0^2)(1 - 1)(0) / (0^4)(1 - 1)^3 = 0/0
We can continue to differentiate and use L'Hospital's rule until we reach a non-indeterminate form. However, in this case, as we can see from the derivatives, the limit does not exist as c approaches 0+.
Therefore, we cannot conclude any specific value for lim c -> 0+ (v).
So, none of the provided options are applicable.